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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :
P1: bringing it in contact with a reservoir at 600 K.
Calculate the change in the entropy (in J/K) of the universe in process P1.
P1: bringing it in contact with a reservoir at 600 K.
Calculate the change in the entropy (in J/K) of the universe in process P1.
Correct answer is '1000J/K'. Can you explain this answer?
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of tempe...
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of tempe...
Change in Entropy of the Universe in Process P1
Given:
- Mass of liquid = 1 kg
- Specific heat of the liquid = 2000 J/Kg-K
- Initial temperature (Ti) = 300 K
- Final temperature (Tf) = 600 K
Concept:
The change in entropy of the universe can be determined using the equation:
ΔS_universe = ΔS_system + ΔS_surroundings
Where:
- ΔS_universe is the change in entropy of the universe
- ΔS_system is the change in entropy of the system
- ΔS_surroundings is the change in entropy of the surroundings
Calculation of ΔS_system:
The change in entropy of the system can be calculated using the equation:
ΔS_system = mcΔT
Where:
- m is the mass of the liquid
- c is the specific heat of the liquid
- ΔT is the change in temperature
Substituting the given values:
ΔS_system = (1 kg)(2000 J/Kg-K)(600 K - 300 K)
= 1 kg * 2000 J/K * 300 K
= 600,000 J/K
Calculation of ΔS_surroundings:
Since the liquid is brought in contact with a reservoir at 600 K, the surroundings are at a constant temperature. Therefore, the change in entropy of the surroundings can be calculated using the equation:
ΔS_surroundings = -Q/T
Where:
- Q is the heat transferred to the surroundings
- T is the temperature of the surroundings
In this case, the heat transferred to the surroundings is equal to the heat absorbed by the liquid. The heat absorbed by the liquid can be calculated using the equation:
Q = mcΔT
Substituting the given values:
Q = (1 kg)(2000 J/Kg-K)(600 K - 300 K)
= 1 kg * 2000 J/K * 300 K
= 600,000 J
Substituting the values of Q and T into the equation for ΔS_surroundings:
ΔS_surroundings = -600,000 J / 600 K
= -1000 J/K
Calculation of ΔS_universe:
Using the equation ΔS_universe = ΔS_system + ΔS_surroundings:
ΔS_universe = 600,000 J/K + (-1000 J/K)
= 599,000 J/K
Therefore, the change in entropy of the universe in process P1 is 599,000 J/K.
Given:
- Mass of liquid = 1 kg
- Specific heat of the liquid = 2000 J/Kg-K
- Initial temperature (Ti) = 300 K
- Final temperature (Tf) = 600 K
Concept:
The change in entropy of the universe can be determined using the equation:
ΔS_universe = ΔS_system + ΔS_surroundings
Where:
- ΔS_universe is the change in entropy of the universe
- ΔS_system is the change in entropy of the system
- ΔS_surroundings is the change in entropy of the surroundings
Calculation of ΔS_system:
The change in entropy of the system can be calculated using the equation:
ΔS_system = mcΔT
Where:
- m is the mass of the liquid
- c is the specific heat of the liquid
- ΔT is the change in temperature
Substituting the given values:
ΔS_system = (1 kg)(2000 J/Kg-K)(600 K - 300 K)
= 1 kg * 2000 J/K * 300 K
= 600,000 J/K
Calculation of ΔS_surroundings:
Since the liquid is brought in contact with a reservoir at 600 K, the surroundings are at a constant temperature. Therefore, the change in entropy of the surroundings can be calculated using the equation:
ΔS_surroundings = -Q/T
Where:
- Q is the heat transferred to the surroundings
- T is the temperature of the surroundings
In this case, the heat transferred to the surroundings is equal to the heat absorbed by the liquid. The heat absorbed by the liquid can be calculated using the equation:
Q = mcΔT
Substituting the given values:
Q = (1 kg)(2000 J/Kg-K)(600 K - 300 K)
= 1 kg * 2000 J/K * 300 K
= 600,000 J
Substituting the values of Q and T into the equation for ΔS_surroundings:
ΔS_surroundings = -600,000 J / 600 K
= -1000 J/K
Calculation of ΔS_universe:
Using the equation ΔS_universe = ΔS_system + ΔS_surroundings:
ΔS_universe = 600,000 J/K + (-1000 J/K)
= 599,000 J/K
Therefore, the change in entropy of the universe in process P1 is 599,000 J/K.
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer?
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer?.
1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer?.
Solutions for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer?, a detailed solution for 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? has been provided alongside types of 1 Kg of a liquid (specific heat = 2000 JK-1kg-1), independent of temperature, is heated from 300 K to 600 K by either of the following process P1 :P1: bringing it in contact with a reservoir at 600 K.Calculate the change in the entropy (in J/K) of the universe in process P1.Correct answer is '1000J/K'. Can you explain this answer? theory, EduRev gives you an
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